A pool is being filled. The following function gives the pool's water level (in centimeters) after $t$ seconds: $W(t)=4\sqrt t+\text{ln}(t+1)$ What is the instantaneous rate of change of the pool's water level after $100$ seconds? Choose 1 answer: Choose 1 answer: (Choice A) A $0.21$ centimeters (Choice B) B $0.21$ centimeters per second (Choice C) C $44.6$ centimeter (Choice D) D $44.6$ centimeters per second
Solution: Understanding the problem The function that represents the instantaneous rate of change of $W(t)$ is its derivative, $W'(t)$. Therefore, the instantaneous rate of change of the water level after $100$ seconds is $W'(100)$. Let's find $W'(t)$ and evaluate it at $t=100$. Finding $W'(t)$ $W'(t)=\dfrac{2}{\sqrt t}+\dfrac{1}{t+1}$ Finding $W'(100)$ $\begin{aligned} W'({100})&=\dfrac{2}{\sqrt {100}}+\dfrac{1}{({100})+1} \\\\ &=\dfrac{2}{10}+\dfrac{1}{101} \\\\ &\approx 0.21 \end{aligned}$ Interpreting units $W(t)$ is the water level in ${\text{centimeters}}$ after $t$ ${\text{seconds}}$. Therefore, we measure its rate of change in ${\text{centimeters}}$ per ${\text{second}}$. In conclusion, the instantaneous rate of change of the pool's water level after $100$ seconds is $0.21$ centimeters per second. The rate of change is positive because the water level is increasing.